这个例子减去了两个列表:

# List of pairs of points
list = []
list.append([(602, 336), (624, 365)])
list.append([(635, 336), (654, 365)])
list.append([(642, 342), (648, 358)])
list.append([(644, 344), (646, 356)])
list.append([(653, 337), (671, 365)])
list.append([(728, 13), (739, 32)])
list.append([(756, 59), (767, 79)])

itens_to_remove = []
itens_to_remove.append([(642, 342), (648, 358)])
itens_to_remove.append([(644, 344), (646, 356)])

print("Initial List Size: ", len(list))

for a in itens_to_remove:
    for b in list:
        if a == b :
            list.remove(b)

print("Final List Size: ", len(list))

在set中查找值比在list中查找值更快:

[item for item in x if item not in set(y)]

我相信这将会比:

[item for item in x if item not in y]

两者都保持了列表的顺序。

from collections import Counter

y = Counter(y)
x = Counter(x)

print(list(x-y))

试试这个。

def subtract_lists(a, b):
    """ Subtracts two lists. Throws ValueError if b contains items not in a """
    # Terminate if b is empty, otherwise remove b[0] from a and recurse
    return a if len(b) == 0 else [a[:i] + subtract_lists(a[i+1:], b[1:]) 
                                  for i in [a.index(b[0])]][0]

>>> x = [1,2,3,4,5,6,7,8,9,0]
>>> y = [1,3,5,7,9]
>>> subtract_lists(x,y)
[2, 4, 6, 8, 0]
>>> x = [1,2,3,4,5,6,7,8,9,0,9]
>>> subtract_lists(x,y)
[2, 4, 6, 8, 0, 9]     #9 is only deleted once
>>>

使用集合差

>>> z = list(set(x) - set(y))
>>> z
[0, 8, 2, 4, 6]

或者你可以让x和y是集合所以你不需要做任何转换。

我们也可以使用set方法来查找两个列表之间的差异

x = [1, 2, 3, 4, 5, 6, 7, 8, 9, 0]
y = [1, 3, 5, 7, 9]
list(set(x).difference(y))
[0, 2, 4, 6, 8]